If $S=\{1,2,3\}$ and $P(\{1,2\}) = 1/3$ and $P(\{2,3\})=2/3$ , what is $P(\{1\}),P(\{2\}),P(\{3\})$?
This can be shown in a linear system as follows (ignoring the inner set notation):
$$P(1) + P(2) + P(3) = 1$$
$$P(1)+P(2) = 1/3$$
$$P(2) + P(3) = 2/3$$
Solving the system gives us that $P(3)=2/3, P(2) =0, P(1) = 1/3$
But $P(2) = 0$ would not make it part of the sample space since it is not an outcome. Where have I gone wrong here?
We call an outcome impossible when it is not an element in the sample space. Impossible outcomes are considered to have a probability of zero.
However elements that are in the sample space may still have a probability measure of zero. These are possible outcomes that are just considered too improbable to give a positive measure.
Having $\mathsf P(\{2\})=0$ when $2$ is an element of the sample space is not a problem.
And you have not gone wrong. That is the answer.